How do you find the integral of int (1 + cos x)^2 dx(1+cosx)2dx?

2 Answers
Mar 8, 2018

int(1+cosx)^2dx=1/4(6x+8sinx+sin2x)+"c"(1+cosx)2dx=14(6x+8sinx+sin2x)+c

Explanation:

First expand the integrand using the perfect square formula

int(1+cos^2x) dx=int1+2cosx+cos^2xdx(1+cos2x)dx=1+2cosx+cos2xdx

Then use the reduction formula on cos^2xcos2x to get it into an integrable form

int1+2cosx+cos^2xdx=int1+2cosx+1/2+1/2cos2xdx=int3/2+2cosx+1/2cos2xdx1+2cosx+cos2xdx=1+2cosx+12+12cos2xdx=32+2cosx+12cos2xdx

Integrate each term using the power rule or standard integrals

int3/2+2cosx+1/2cos2x=3/2x+2sinx+1/4sin2x+"c"=1/4(6x+8sinx+sin2x)+"c"32+2cosx+12cos2x=32x+2sinx+14sin2x+c=14(6x+8sinx+sin2x)+c

Mar 8, 2018

int(1+cos x)^2 dx(1+cosx)2dx
=int(1+2cos x +cos^2 x) dx(1+2cosx+cos2x)dx
=int(1+2cos x +(1/2cos 2x - 1/2) dx(1+2cosx+(12cos2x12)dx *
=int(1/2 + 2cos x + 1/2cos 2x) dx(12+2cosx+12cos2x)dx
=1/2x +2sin x +1/4sin 2x +c12x+2sinx+14sin2x+c

*1 + 2cos^2 x = cos 2x1+2cos2x=cos2x
:. cos^2 x = 1/2cos 2x -1/2